If a number N is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27? A. 1 B. 2 C. 3 D. 4 E. 6 The OA is the option C. Experts, how can I show that C is the correct answer? May you help me here? I'd be thankful.

VJesus12 wrote:If a number N is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27? A. 1 B. 2 C. 3 D. 4 E. 6 The OA is the option C. Experts, how can I show that C is the correct answer? May you help me here? I'd be thankful. For N to be divisible by 33 means N/(3*11) = some integer, call it X. So N = 3*11*X. Raising N to the k means: N^k = 3^k*11^k*X^k. Dividing this by 27 means dividing by 3^3, so N^k/(3^3) = (3^k*11^k*X^k)/3^3 Now, there could be other powers of 3 hiding in the X depending on what numbers are actually used, but to guarantee that 3^3 can divide into the numerator, 3^k has to be at least as great as 3^3, meaning that k = 3, C

N is divisible by 33, so N = 33*f (where f is an unknown factor) = 3*11*f N^k is divisible by 27, so N^k = 27*g (where g is an unknown factor) = 3*3*3*g = (3^3)*g Combining these 2 facts, we get [3*11*f]^k = (3^3)*g 3^k * (11*f)^k = (3^3)*g Comparing similar bases (namely powers of 3 in this case), we see that 3^k = 3^3 (minimum) Comparing powers we see that k = 3 also ANSWER = C

VJesus12 wrote:If a number N is divisible by 33, what will be the minimum value of k, such that N^k should definitely be divisible by 27? A. 1 B. 2 C. 3 D. 4 E. 6 Since N is divisible by 33, we see that N has, at a minimum, one prime factor of 3. In order for N^k/27 = integer, we see that N must have at least three primes of 3 (since 27 has three factors of 3), and thus the minimum value of k would have to be 3, since 33^3 = 3^3 x 11^3 = 27 x 11^3. Answer: C

If a number N is divisible by 33, what will be the minimum - The Beat The GMAT Forum

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